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Given 3 diophantine equations:

$$x_1y_1+x_2y_2=x_3y_3+x_4y_4$$

and

$$x_1+x_2 = x_3+x_4$$ and

$$y_1+y_2 = y_3+y_4$$

We're interested in solutions to this system of equations when all variables are positive. I conjecture that any solutions have $x_1=x_3$ or $x_1=x_4$ and $y_1=y_3$ or $y_1=y_4$. Any tips on how to go about proving this? Thanks.

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1 Answer 1

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No you are wrong. For any $x_1,x_2,x_3,x_4$ let $y_1=x_3, y_2=x_4,y_3=x_1, y_4=x_2$ and you get a solution with all $x$ distinct and all $y$ distinct.

If you want all $8$ numbers distinct, here is an example:

$(x_1,x_2,x_3,x_4)=(5,1,4,2)$

$(y_1,y_2,y_3,y_4)=(11,9,12,8)$

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  • $\begingroup$ Ah, that was fast. Thanks. Did you just use trial and error to get the solution $\endgroup$
    – Ameet Sharma
    Nov 11, 2015 at 4:10
  • $\begingroup$ Well yes but not fully. Essentially you can show that the system is equivalent to the last two equations combined with $(x_1-x_2)(y_1-y_2)=(x_3-x_4)(y_3-y_4)$ and you can get a lot of non-trivial solutions from here. $\endgroup$
    – cr001
    Nov 11, 2015 at 4:21

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